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$\begingroup$

I managed to get
$$\sum_{k=0}^n \binom{n+1}{k+1}(-1)^k$$ on the left side, but I don't know how to proceed from here. thanks in advance.

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    Duplicate: [How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$?](http://math.stackexchange.com/q/38623) Found [using Approach0](https://approach0.xyz/search/?q=%24%5Csum%5Climits_%7Bk%3D0%7D%5En%20%5Cbinom%7Bn%7D%7Bk%7D%20%5Cfrac%7B(-1)%5Ek%7D%7Bk%2B1%7D%20%3D%20%5Cfrac%7B1%7D%7Bn%2B1%7D%24&p=1). – 2017-01-07

2 Answers 2

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Expand $(1-1)^{n+1}$ by using the Binomial Theorem: $$0=(1-1)^{n+1}=1-\sum_{k=0}^{n}\binom{n+1}{k+1}(-1)^{k}=1-(n+1)\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{k+1}.$$

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    Thanks alot! but I'm not sure how you get the "1-" before the sum? thanks again! – 2017-01-07
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    @Omer.S Because $(1-1)^{n+1}=\binom{n+1}{0}+\sum_{k=0}^{n}\binom{n+1}{k+1}(-1)^{k+1}$. – 2017-01-07
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    I understand that $\ (1-1)^{n+1} = \sum_{k=0}^{n+1} {n+1\choose k+1} (-1)^{k+1}$ but why $\ {n+1\choose 0}$? – 2017-01-07
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    @Omer.S $$(1-1)^{n+1}=\sum_{j=0}^{n+1}\binom{n+1}{j}(-1)^{j}=\binom{n+1}{0}+\sum_{j=1}^{n+1}\binom{n+1}{j}(-1)^{j}\\=\binom{n+1}{0}+\sum_{k=0}^{n}\binom{n+1}{k+1}(-1)^{k+1}.$$ – 2017-01-07
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$$\begin{eqnarray*}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k+1}&=&\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx\\&=&\int_{0}^{1}(1-x)^n\,dx\\&=&\int_{0}^{1}z^n\,dz = \color{red}{\frac{1}{n+1}}.\end{eqnarray*}$$